Solved Problems on Armature of a Dc Machine Continuation of Chapter 2-VK

Solved Problems on Armature of a DC machine Example: 1 Determine the number of poles, armature diameter and core length for the preliminary design of a 500kW, 400V, 600 rpm, dc shunt generator assuming an average flux density in the air gap of 0.7 T and specific electric loading of 38400 ampere- conductors per metre. Assume core length/ pole arc = 1.1. Apply suitable checks. Output in watts 500×103 Armature current Ia = IL + ISh = = =400A terminal voltage 400 In order that the current/path is not more than 200A, a lap winding is to be used. The number of parallel paths required is A=

1250 = 6.25 200

Since the number of parallel paths cannot be a fraction as well an odd integer, A can be 6 or 8. Let it be 6. (Note: A current/path of 200A need not strictly be adhered) Check: Frequency of the induced emf f =

PN 6×60 = =30Hz and can be considered 120 120

as the frequency generally lies between 25 and 50 Hz D L =

kW 1.64 × 10 B q N

=

1.64 ×

10

500 × 0.7 × 38400 × 600

≈ 0.189m Note: Since the capacity is considerable, power developed in the armature can be taken as power output for a preliminary design. L = 1.1 pole arc=1.1 ψ τ Since ψ lies between 0.6 and 0.7, let it be 0.7.

1

Since τ=

πD πD , L=1.1 ×0.7× =0.4D 6 P

D × 0.4D = 0.189 D =

0.189 = 0.473m 0.4

D = 0.78m and L = 0.4 × 0.78 ≈ 0.31m πDN π×0.78×600 = =24.2m/s 60 60 and is within the limits,i.e.30 m/s. Check: Peripheral velocity v=

Example: 2 Determine the main dimensions of the armature core, number of conductors, and commutator segments for a 350kW, 500V, 450 rpm, 6pole shunt generator assuming a square pole face with pole arc 70% of the pole pitch. Assume the mean flux density to be 0.7T and ampereconductors per cm to be 280. D L = =

kW 1.64 × 10 B q N

1.64 ×

10

350 × 0.7 × 280 × 100 × 450 ≈ 0.24m

For a square pole face, L= ψτ =0.7× πD/6=0.36D D3 =

0.24 =0.654m3 0.37

Therefore D=0.87m and L=0.37×0.87≈0.32m Since E= φZNP/60A, number of armature conductors Z=E60A/φNP

2

Flux per pole =

Bav πDL 0.7×π×0.87×0.32 = =0.1Wb P 6

If a lap winding is assumed, A=P and therefore Z=

60× 500× 6 =666.6 and is not possible as the number of conductors must always an even integer. 0.1× 450× 6

The number of conductors can be 666 or 668. Let it be 666. Number of coils=Number of commutator segments =

Z 666 = =333 if asingle turn coil is assumed. 2 ×turns per coil 2×1

Example: 3 A design is required for a 50kW,4pole,600rpm, and 220V dc shunt generator. The average flux density in the air gap and specific electric loading are respectively 0.57T and 30000 ampere- conductors per metre. Calculate suitable dimensions of armature core to lead to a square pole face. Assume that full load armature drop is 3% of the rated voltage and the field current is 1% of rated full load current. Ratio pole arc to pole pitch is 0.67.

D L =

kW 1.64 × 10 B q N

Note: Since data is available in the problem to calculate the power developed in the armature, the same may be obtained to substitute for kW. Power developed in the armature in kW=EIa ×10-3 E=V+voltage drop in the armature circuit=220+0.03×220=226.6V Ia = IL + ISh IL =

Output in watts 50 × 10 = = 227.3A terminal voltage 220

Ia = 227.3 + 0.01×227.3=229.6A kW=226.6 ×229.6×10-3 =52

3

D L =

1.64 ×

10

52 × 0.57 × 300 × 600 ≈ 0.031m

For a square pole face, L= ψ τ =0.67× πD/4=0.53D D3 =

0.031 3 m and therefore D=0.39m and L=0.53×0.39≈0.21m 0.53

Example: 4 Determine the diameter and length of the armature core for a 55kW, 110V, 1000rpm, and 4pole dc shunt generator. Assume: Specific magnetic loading 0.5T, Specific electric loading 13000 ampere –turns, Pole arc 70% of pole pitch and length of core about 1.1 times the pole arc, Allow 10A for field current and a voltage drop of 4V for the armature circuit. Determine also the number of armature conductors and slots. D L =

kW 1.64 × 10 B q N

Power developed in the armature in kW = EIa × 10 E=V+ voltage drop in the armature circuit=110+4=114V Ia = IL + ISh =

55 × 10 + 10 = 510A 110

kW=114 ×510×10-3 =58.14 q = 13000 × 2 = 26000 as q must be substituted in ampere conductors andone turn corresponds 2 conductors D L =

1.64 ×

10

58.14 × 0.5 × 26000 × 1000 ≈ 0.027m

since L=1.1 times pole arc, L= ψ τ =1.1×0.7× πD/4=0.4D D3 =

0.027 3 m and therefore D=0.36m and L=0.6×0.36≈0.22m 0.6

4

Number of armature conductors Z=

60EA φNP

Flux per pole φ=

Bav πDL 0.5×π×0.36×0.22 = =0.03Wb P 4

If a lap winding is assumed, A=P and therefore Z=

60× 114× 4 ≈ 228 0.03× 1000× 4

Since the minimum number of slots per pole is 9 from commutation point of view, minimum number of slots for the machine is = 9 x 4 = 36 -------(1) Since slot pitch λs = lies between

πD , lies between 2.5 and 3.5 cm, the number of slots S

π × 36 π × 36 ≈32 and ≈ 45 -------------------(2) 3.5 2.5

From 1 and 2 the number of slots lies between 36 and 45. Since a lap winding is assumed and for a lap winding the number of slots may be multiple of number of poles i.e. 6 or multiple of number of pair poles i.e. 3, let the number of slots S = 40. Note: Since preliminary number of slots and conductors are known, actual number of conductors per slot and number of conductors can be calculated as detailed below. Conductors per slot=

228 =5.7and is not possible as Conductors per slot must always be 40

an even integer for a double layer winding. Since only double layer winding is for dc machines, let the number Conductors per slot be 6. Therefore Zrevised =40×6=240. Example: 5 For a preliminary design of a 50hp, 230V, 1400 rpm dc motor, calculate the armature diameter and core length, number of poles and peripheral speed. Assume specific magnetic loading 0.5T, specific electric loading 25000 ampere- conductors per meter, efficiency 0.9. Ia = IL - ISh

Input in watts ℎ8 × 746:9 50 × 747 ≈ IL = = = =180.2A Voltage ; 0.9 × 230

For this armature current, a lap or wave winding can be used. Since minimum number

5

of paths

and poles is two, 2 poles are sufficient for the machine. However, to gain more

advantages of more number of poles, let P=4. 4 × 1400 = = 46.7?@, within the limits. 120 120

D2 L=

kW 1.64×10-4 Bav q N

Power developed in the armature 1 + 29 =A B outputpower 39 =A

1 + 2 × 0.9 B 50 × 0.746 = 38.7kW 3 × 0.9

D2 L=

Since

38.7 1.64×10 ×0.5×25000×1400 -4

C lies between 0.55 and 1.1, let it be 1.0. D

Therefore L=τ= I =

= 0.0134m3

HI = 0.785I 4

0.0134 = 0.017m3 and 0.785

Peripheral velocity v =

D=0.26m,

L=0.785×0.26≈0.2m

HI> H × 0.26 × 1400 = ≈ 19m/s 60 60

Example: 6 Calculate the armature diameter and core length for a 7.5kW, 4pole, 1000rpm, and 220V shunt motor. Assume: Full load efficiency = 0.83, field current is 2.5% of rated current. The maximum efficiency occurs at full load. D2 L=

kW 1.64×10-4 Bav q N

Power developed in the armature 6

= Electrical input to the motor − Mield and armature copper losses = Electrical output of the motor + Iron, friction and windage losses output 7.5×103 Electrical input to the motor = = ≈ 9036W η 0.83 Losses at full load =Q

RS S

T output = Q

RU.V U.V

T × 7.5 × 103 = 1536W

Since efficiency is maximum at full load and at maximum efficiency, the variable loss is equal to constant loss, Variable loss = armature copper loss = Field copper loss =XYZ[ℎ X\2 ][ℎ ^_ ;Z[ℎ

1536 ≈ 768W 2

7.5×103 Ish =0.025 full load current =0.025× = 1.03A 0.83×220

Field copper loss =220×1.03=226.6W

Power developed in the armature = 9036 − 768 − 226.6 = 8041.4W OR Iron, friction and windage losses = Constant losses − Mield copper loss = 768 − 226.6 = 541.4W OR Power developed in the armature = 7500 + 541.4 = 8041.4W Since specific magnetic loading lies between 0.45 and 0.75 T, let it be 0.6T. Since specific electric loading lies between 15000 to 50000 ampere-conductors, let it be 30000. D2 L= Since

8.041 1.64×10-4 ×0.6×30000×1000

C lies between 0.55 and 1.1, let it be 1.0. D

Therefore L=τ= I =

= 0.0027m3

HI = 0.785I 4

0.0027 = 0.0343m3 and 0.785

D=0.33m,

L=0.785×0.33≈0.26m 7

Example: 7 List the design details of armature winding suitable for a 35hp, 4pole, 500rpm, and 230V, motor. Flux per pole 0.028Wb. Details of the winding : Type of winding, number of slots, number of conductors, crosssectional area of conductor, number of coils, back pitch, front pitch etc. Input in watts ℎ8 × 746:9 Ia ≈ = Voltage ; 35 × 747 = =126.1A with the assumption, efMiciency is 0.9. 0.9 × 230 For the armature current of 126.1A, a lap or wave winding can be used. Let a wave winding be used. Number of armature conductors Z=

60EA 60×230×2 = ≈ 492 φNP 0.028×500×4

Since the minimum number of slots per pole is 9 from commutation point of view, minimum number of slots for the machine is 9 x 4 = 36. [Note: Since the diameter of the armature is not known, number of slots from the slot pitch consideration cannot be fixed. It is not very much essential to determine D either from D L product or peripheral velocity and then fix number of the slots. Even the condition minimum, 9 slots per pole need not be satisfied, as it is only a guiding figure in fixing the number of slots for the machine. For a preliminary design, a number, around the minimum number of slots can be selected.] Since for a wave winding, the number of slots should not be a multiple of pair of poles i.e. p = 2, let the number of slots be 43. Conductors per slot=

492 ≈11.5 and is not possible. Let the number Conductors per slot be 12. 43 Therefore Zrevised =12×43 =516

Commutator or average pitch YC for simplex wave winding=

C±1 p

and must be an integer Number of coils C=

Z 516 = = 258 with single turn coils Yassumed\. 2×turns per coil 2×1

8

C = 258 leads to an asymmetrical, unbalanced or a wave winding with a dummy coil because 258 is a multiple of number of pair of poles. If a two turn coil is assumed, then the number coils will be

516

2×2

= 129 and leads to a

symmetrical, balanced or a wave winding without a dummy coil because 129 is not a multiple of number of pair of poles. Therefore YC =

C±1 129±1 = = 65 or 64 p 2

If YC = 65, YB = 65 and YF = 65 If YC =64, YB = YC ±1 = 64 ±1 = 63 or 65 With YB = 63, YF

= YB ± 2 = 63 ± 2 = 65 so that

With YB = 65, YF

= YB ± 2 = 65 ± 2 = 63 so that I

Cross-sectional area of the conductor a = Aaδ =

126.1 2×5

YB + YF = YC = 64 2 YB + YF = YC = 64 2

=12.61mm2

For the above area a round or rectangular conductor can be used. Example: 8 For a preliminary design of a 1500kW, 275V, 300rpm, dc shunt generator determine the number of poles, armature diameter and core length, number of slots and number of conductors per slot. Assume: Average flux density over the pole arc as 0.85T, Output coefficient 276, Efficiency 0.91.Slot loading should not exceed 1500A. Ia = IL + ISh ≈ IL =

1500 × 10 = 5454.5A 275

In order that the current per path is not more than about 200A,a simplex wave winding cannot be used. Obviously, a lap winding has to be used having number of parallel paths A=

5454.5 ≈ 27.27. 200

9

Since the number of parallel paths must be an even integer, it can be 26 or 28. Let A = 28.Therefore, with a simplex lap winding considered P = 28. hi

Check: f = = R U

V × UU R U

= 70Hz and is on the higher side as frequency generally considered

between 25 and 50Hz. In order to reduce the frequency and to have A =28, a duplex lap winding can used with P = 14 and f = 35Hz.

D L =

kW kW =  1.64 × 10 B q N Co N

Power developed in the armature in kW =

output 1500 = = 1648.1 9 0.91

[Note: a\ When the speed is in rpm in the expression, D L =

kW , the output coefMicient Co N

Co = 1.64 × 10 B q lies between

1.64 × 10 × 0.45 × 15000 = 1.0 and 1.64 × 10 × 0.75 × 50000 = 6.0.

On the other hand if the speed is in rpm, Co = H × 10 B q lies between H × 10 × 0.45 × 15000 = 60 and H × 10 × 0.75 × 50000 = 360.

Since the given value of Co lies in the range of 60 and 360, the speed must be in rps when substituted in the output equation.]

D L =

1648.4

276 ×

UU lU

≈ 1.2m

[Note: In order to spilt up D L product into D and L, a value for

has to be assumed. ]

Let L=τ, Therefore, L= D3 =

n o

ratio or peripheral velocity

πD =0.23D 14

1.2 = 5.2m3 and therefore D=1.7m and L=0.23×1.7≈0.39m 0.23

Since the minimum number of slots per pole is 9 from commutation point of view, minimum number of slots for the machine is = 9 x 44 =126 -------(1)

10

Since slot pitch λp = lies between

HI , lies between 2.5 and 3.5 cm, the number of slots q

π × 170 π × 170 ≈152 and ≈ 214 -------------------(2) 3.5 2.5

From 1 and 2 the number of slots lies between 152 and 214.

Since a lap winding is assumed and for a lap winding the number of slots may be multiple of number of poles i.e. 14 or multiple of number of pair poles i.e. 7, let the number of slotsS = 196. Number of armature conductors Z=

E60A φNP

Bav πDL ψBg πDL = P P 0.7×0.85×π×1.7×0.39 = =0.089Wb with the assumption Ψ= 0.7 4

Flux per pole φ=

Z=

60× 275× 28 ≈ 1236 0.089× 300×14

Conductors per slot=

1236 =6.3and is not possible as Conductors per slot must always be 196

an even integer for a double layer winding. Let the number Conductors per slot be 6. Therefore Zrevised =196×6=1176. Slot loading = conductors per slot × current through the conductor i.e. =6×

5454.5 28

Ia A

=1168.8 < 1500 and the condition is satisMied.

Example: 9 A 150hp, 500V, 6pole, 450rpm, dc shunt motor has the following data. Armature diameter = 54cm, length of armature core = 24.5cm, average flux density in the air gap = 0.55T, number of ducts = 2, width of each duct = 1.0cm, stacking factor = 0.92. Obtain the number of armature slots and work the details of a suitable armature winding. Also determine the dimensions of the slot. The flux density in the tooth at one third height from the root should not exceed 2.1T.

11

Ia =

hp×746 150×746 = =248.7A η× V 0.9×500

For this armature current, a lap or wave winding may be used. Let a lap winding be used. Since the minimum number of slots per pole is 9 from commutation point of view, minimum number of slots for the machine is = 9 x 6 = 54 -------(1) Since slot pitch λs = lies between

πD , lies between 2.5 and 3.5 cm, the number of slots S

H × 54 H × 54 ≈48 and ≈ 68 -------------------(2) 3.5 2.5

From 1 and 2 the number of slots lies between 54 and 68.

Since a lap winding is assumed and for a lap winding the number of slots may be multiple of number of poles i.e. 6 or multiple of number of pair poles i.e. 3, let the number of slots S = 60. Details of winding: Type of winding (already fixed), number of slots (also fixed), number of conductors, cross-sectional area of the conductor, back pitch, front pitch etc. Flux per pole φ=

Bav πDL 0.55×π×0.54×0.245 = =0.038Wb P 6

qince a lap winding is assumed, Conductors per slot=

Z=

60×500× 6 ≈ 1754 0.038× 450× 6

1754 =29.2and is not possible 60

let the number Conductors per slot be 30. Therefore Zrevised =30×60=1800. Z 1800 1800 Back pitch YB = ±K = ±K = ± 1 = 301 say. P 6 6 Front pitch YF =YB ±2 = 301±2=299 for a progressive winding Since the current density lies between 4.5 and 7A/mm2, let it be 5A/mm2 Cross-sectional area of the conductor a =

Zt 248.7 = = 8.3mm uv 6 × 5

Since a is less than 10mm , let a round conductor be used of bare diameter

12

4x 4 × 8.3 w =w = 3.25mm H H Since only double layer winding is used for a dc machine, number of conductor per layer is (30/2) =15 and can be arranged in any one of the following ways.

(A)- 15 conductors are arranged one below the other in each layer. Slot is not proportionate. (B)- 15 conductors are arranged one beside the other in each layer. Slot is not proportionate. (C) and ( D) - arrangement of conductors as shown in (C) or (D), leads to proportionate slots If the conductors are arranged as shown in (D), then,

Slot width bs= ( diameter of the conductor + insulation on it) number of conductors along the

13

slot width + (insulation over the coil side or group of coil sides + slot liner + clearance) = (3.25 + 2 x 0.075)3 + 1.5 = 11.7mm Slot depth ht= (diameter of the conductor + insulation on it) number of conductors along the slot depth + (insulation over the coil side or group of coil sides + slot liner + separator + clearance) + wedge 3 to 5 mm + lip 1 or 2mm. = (3.25 + 2 x 0.075)10 +4 + 4 + 2 = 44mm Flux density in the tooth at one third height from the root of the tooth yz{: = width of the tooth at 1:3 height from the root of the tooth ~z{: | =

H QI − q

‚ƒ 

T

− ~p =

H Q54 −

60

×. 

T

|

}

€

~z{: × C ×  |

− 1.17 = 1.35cm

Net iron length C = „ (L -…† ~† )=0.9(24.5- 2×1)= 20.7cm ‡ℎˆ_ˆI C 1.64 × 10 × 26906 × 600 × 0.75 × 0.3

II machine:

350: Š‹ 0.91 I C= = ≈ 0.195   1.64 × 10 yt† ‰> 1.64 × 10 × 0.61 × 26906 × 720

For a square pole face, L= ψ τ =0.66× πD/6=0.35D D3 =

0.195 =0.564m3 0.35

Therefore D=0.83m and L=0.35×0.83≈0.29m Since the minimum number of slots per pole is 9 from commutation point of view, minimum number of slots for the machine is = 9 x 6 = 54 -------(1)

Since slot pitch lies between

λs =

πD S , lies between 2.5 and 3.5 cm, the number of slots

H × 83 H × 83 ≈74 and ≈ 104 -------------------(2) 3.5 2.5

From 1 and 2 the number of slots lies between 74 and 104.

If a lap winding is assumed then for a lap winding the number of slots may be multiple of number of poles i.e. 6 or multiple of number of pair poles i.e. 3, let the number of slots S = 78 Number of armature conductors Z=

E60A φNP

. emf induced E = 440+0.04×440=457.6V Flux per pole φ= Z=

Bav πDL 0.61×π×0.83×0.29 = =0.078Wb P 6

60× 457.6×6 ≈ 488 0.078× 720× 6

Conductors per slot=

488 =6. 2 and is not possible. Let it be 6 78 16

Zrevised =78×6=468

Conductors /layer=

6 =3 3

Three conductors /layer is possible only with 3 coil sides/layer of single turn coil or one coil Side/layer of 3 turn coil. If a single turn coil is assumed, then the number of coils =

lV

=234.

There the number of segments = 234. Check: Commutator segment pitchτC =

πDC number of segments

Commutator diameter IŒ =0.7×83=58.1cm DŒ =

H × 58.1 = 0.78m > 4mm 234

Voltage between segments pitch =

Open circuit voltage number of

segments pole

=

457.6 =11.7V

–ℎˆ_ˆx = …’m~ˆ_^